Using pointers in C

Question

#include <stdio.h>

int main(void)
{
    int a[5]={1,2,3,4,5};
    int *ptr=(int*)(&a+1);
    printf("%d %d\n",*(a+1),*(ptr-1));
    return 0;
}

Output:

2,5

I could not understand how the *(ptr-1) evaluated to 5 (correct output). But when I manually did it was 1. My understanding is *(ptr-1) will get evaluated to *(&a+1-1) which would be *(&a) which is 1.

Please help me to understand this concept.

&a has type pointer to array of 5 ints, not pointer to int. Your cast here is hiding your mistake. — τεκ
– τεκ, Commented Aug 25, 2011 at 17:11
What is 'yashwant kanetkar'? How is it relevant to the question? — Jonathan Leffler
– Jonathan Leffler, Commented Aug 25, 2011 at 18:08

phoxis · Accepted Answer · 2011-08-25 17:21:20Z

8

int *ptr=(int*)(&a+1);

makes &a + 1 makes &a + sizeof (a) as a is type of int [5] which makes ptr to point to actually a[5] (invalid/beyond the defined limit)

(ptr - 1) points thus points to a[4] and *(ptr - 1) will print 5.

answered Aug 25, 2011 at 17:21

phoxis

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Comments

user195488 · Accepted Answer · 2011-08-25 17:11:25Z

5

In your printf, you are retrieving the value in pos 1 by incrementing the position with +1 from 0.

In your second integer, you are retrieving 5 because &a+1 actually points beyond the array, so when you do *(ptr-1), it returns 5. If you remove the -1, then you get a really weird result (in my case it was -1078772784).

answered Aug 25, 2011 at 17:11

user195488

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